The most common force is the variable force, which is more commonly encountered. Figure 6.2 is a plot of a varying force in one dimension. If the displacement \(\Delta x\) is small, we can take the force \(F(x)\) as approximately constant and the work done is then
\(\Delta W=F(x) \Delta x\)
This is illustrated in Figure 6.2(a). Adding successive rectangular areas in Figure 6.2(a) we get the total work done as
\(W \cong \sum_{x_{i}}^{x_{f}} F(x) \Delta x\)
where the summation is from the initial position \(x_{i}\) to the final position \(x_{f}\)

If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Figure 6.2(b). Then the work done is
\(
\begin{aligned}
W &=\lim _{\Delta x \rightarrow 0} \sum_{x_{i}}^{x_{f}} F(x) \Delta x \\
&=\int_{x_{i}}^{x_{f}} F(x) \mathrm{d} x \dots (6.6)
\end{aligned}
\)
where ‘lim’ stands for the limit of the sum when \(\Delta x\) tends to zero. Thus, for a varying force the work done can be expressed as a definite integral of force over displacement.

Example 6.8:

A woman pushes a trunk on a railway platform that has a rough surface. She applies a force of \(100 \mathrm{~N}\) over a distance of \(10 \mathrm{~m}\). Thereafter, she gets progressively tired and her applied force reduces linearly with distance to \(50 \mathrm{~N}\). The total distance through which the trunk has been moved is \(20 \mathrm{~m}\). Plot the force applied by the woman and the frictional force, which is \(50 \mathrm{~N}\) versus displacement. Calculate the work done by the two forces over \(20 \mathrm{~m}\).

Solution: The plot of the applied force is shown in the Figure below. At \(x=20 \mathrm{~m}, F=50 \mathrm{~N}(\neq 0)\). We are glven that the frictional force \(f\) is \(|\vec{f}|=50 \mathrm{~N}\). It opposes motion and acts in a direction opposite to \(\vec{F}\). It is, therefore, shown on the negative side of the force axis.
The work done by the woman is
\(W_{F} \rightarrow\) area of the rectangle \(\mathrm{ABCD}+\) area of the trapezium CEID
\(
\begin{aligned}
W_{F} &=100 \times 10+\frac{1}{2}(100+50) \times 10 \\
=& 1000+750 \\
&=1750 \mathrm{~J}
\end{aligned}
\)
The work done by the frictional force is
\(
\begin{array}{c}
W_{f} \rightarrow \text { area of the rectangle AGHI } \\
W_{f}=(-50) \times 20 \\
=-1000 \mathrm{~J}
\end{array}
\)
The area on the negative side of the force axis has a negative sign.

Figure 6.3

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